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12-e7c4a452-0
- 首先我嘗試寫(xiě)了一下這段代碼如下：
private static void testForeachAddOrRemove() {
var list = new ArrayList<String>();
list.add("1");
list.add("2");
for (var s : list) {
if ("2".equals(s)) {
list.remove(s);
}
}
list.forEach(System.out::println);
}
調(diào)用結(jié)果：異常ConcurrentModificationException如下：
12-e7c4a452-1
- 如圖所示提示ArrayList.java 1043行拋出異常，我用的JDK11，馬上點(diǎn)進(jìn)去看了源碼，正是如下這行拋出的異常，原因是modCount != expectedModCount那么這兩個(gè)變量什么意思呢？繼續(xù)讀源碼
final void checkForComodification() {
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
}
- modCount是在抽象類(lèi)AbstractList中被定義，是集合結(jié)構(gòu)修改的次數(shù)，例如添加元素、刪除元素等結(jié)構(gòu)調(diào)整都會(huì)使modCount++,下面列舉add()和刪除的源碼，其他結(jié)構(gòu)化修改也會(huì)modCount++，這里不做列舉。
public boolean remove(Object o) {
final Object[] es = elementData;
final int size = this.size;
int i = 0;
found: {
if (o == null) {
for (; i < size; i++)
if (es[i] == null)
break found;
} else {
for (; i < size; i++)
if (o.equals(es[i]))
break found;
}
return false;
}
fastRemove(es, i);//-----------重點(diǎn)重點(diǎn)重點(diǎn)------------------
return true;
}
private void fastRemove(Object[] es, int i) {
modCount++;//---------------重點(diǎn)重點(diǎn)重點(diǎn)------------------
final int newSize;
if ((newSize = size - 1) > i)
System.arraycopy(es, i + 1, es, i, newSize - i);
es[size = newSize] = null;
}
public boolean add(E e) {
modCount++;//-------重點(diǎn)重點(diǎn)重點(diǎn)----------
add(e, elementData, size);
return true;
}
- expectedModCount是在ArrayList的內(nèi)部類(lèi)Itr中聲明的，Itr實(shí)現(xiàn)類(lèi)迭代器接口，這里做了下精簡(jiǎn)只保留了變量聲明和next()方法、remove()方法，可以看到迭代器的實(shí)現(xiàn)里聲明了expectedModCount = modCount；這里理解成預(yù)期結(jié)構(gòu)化調(diào)整次數(shù) = 結(jié)構(gòu)化調(diào)整次數(shù)，為什么每次迭代next()和remove()之前都要檢查是否相等呢？可以理解成如果沒(méi)有這個(gè)校驗(yàn)，某個(gè)線程刪除了list的一個(gè)元素，此時(shí)next方法不知道size變更了，依然去取數(shù)組里的數(shù)據(jù)，會(huì)導(dǎo)致數(shù)據(jù)為null或ArrayIndexOutOfBoundsException異常等問(wèn)題。
/**
* An optimized version of AbstractList.Itr
*/
private class Itr implements Iterator<E> {
int cursor; // index of next element to return
int lastRet = -1; // index of last element returned; -1 if no such
int expectedModCount = modCount;//-------------重點(diǎn)重點(diǎn)重點(diǎn)---------------
@SuppressWarnings("unchecked")
public E next() {
checkForComodification();//-------------重點(diǎn)重點(diǎn)重點(diǎn)------------------
int i = cursor;
if (i >= size)
throw new NoSuchElementException();
Object[] elementData = ArrayList.this.elementData;
if (i >= elementData.length)
throw new ConcurrentModificationException();
cursor = i + 1;
return (E) elementData[lastRet = i];
}
public void remove() {
if (lastRet < 0)
throw new IllegalStateException();
checkForComodification();//-----------重點(diǎn)重點(diǎn)重點(diǎn)---------------------
try {
ArrayList.this.remove(lastRet);
cursor = lastRet;
lastRet = -1;
expectedModCount = modCount;//------------重點(diǎn)重點(diǎn)重點(diǎn)-------------------
} catch (IndexOutOfBoundsException ex) {
throw new ConcurrentModificationException();
}
}
final void checkForComodification() {
if (modCount != expectedModCount)//-----------重點(diǎn)重點(diǎn)重點(diǎn)---------------------
throw new ConcurrentModificationException();
}
}
- 這里串起來(lái)就好理解了，由于增強(qiáng)for循環(huán)底層反編譯之后是迭代器實(shí)現(xiàn)的，所以在iterator初始化的時(shí)候（也就是for循環(huán)開(kāi)始處），expectedModCount = modCount之后for循環(huán)內(nèi)部進(jìn)行remove實(shí)際上用的是ArrayList的remove()方法，執(zhí)行了modCount++，而進(jìn)行for循環(huán)底層進(jìn)行next()迭代的時(shí)候進(jìn)行了checkForComodification()方法判斷，modCount++了next()并不知道，所以造成不相等拋出異常。
- 那為什么使用迭代器可以進(jìn)行刪除而不拋出異常呢？因?yàn)榭瓷厦娴脑创a，迭代器內(nèi)部的remove()方法的實(shí)現(xiàn)在調(diào)用ArrayList.this.remove()進(jìn)行刪除之后，expectedModCount = modCount;及時(shí)同步了這兩個(gè)變量的值。所以使用迭代器刪除不會(huì)造成問(wèn)題，寫(xiě)法如下：
private static void testForeachAddOrRemove() {
var list = new ArrayList<String>();
list.add("1");
list.add("2");
/*for (var s : list) {
if ("2".equals(s)) {
list.remove(s);
}
}*/
var iterator = list.iterator();
while (iterator.hasNext()) {
var str = iterator.next();
if ("2".equals(str)) {
iterator.remove();
}
}
list.forEach(System.out::println);
}
建議：如果是JDK8或者以上版本，推薦使用removeIf進(jìn)行刪除，這也是IDEA推薦寫(xiě)法
寫(xiě)法如下：
private static void testForeachAddOrRemove() {
var list = new ArrayList<String>();
list.add("1");
list.add("2");
/*for (var s : list) {
if ("2".equals(s)) {
list.remove(s);
}
}*/
/*var iterator = list.iterator();
while (iterator.hasNext()) {
var str = iterator.next();
if ("2".equals(str)) {
iterator.remove();
}
}*/
list.removeIf("2"::equals);
list.forEach(System.out::println);
}
- 同樣可以閱讀一下removeIf()的源碼
@Override
public boolean removeIf(Predicate<? super E> filter) {
return removeIf(filter, 0, size);
}
/**
* Removes all elements satisfying the given predicate, from index
* i (inclusive) to index end (exclusive).
*/
boolean removeIf(Predicate<? super E> filter, int i, final int end) {
Objects.requireNonNull(filter);
int expectedModCount = modCount;//-----------重點(diǎn)重點(diǎn)重點(diǎn)--------------
final Object[] es = elementData;
// Optimize for initial run of survivors
for (; i < end && !filter.test(elementAt(es, i)); i++)
;
// Tolerate predicates that reentrantly access the collection for
// read (but writers still get CME), so traverse once to find
// elements to delete, a second pass to physically expunge.
if (i < end) {
final int beg = i;
final long[] deathRow = nBits(end - beg);
deathRow[0] = 1L; // set bit 0
for (i = beg + 1; i < end; i++)
if (filter.test(elementAt(es, i)))
setBit(deathRow, i - beg);
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
modCount++;//-------------重點(diǎn)重點(diǎn)重點(diǎn)------------
int w = beg;
for (i = beg; i < end; i++)
if (isClear(deathRow, i - beg))
es[w++] = es[i];
shiftTailOverGap(es, w, end);
return true;
} else {
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
return false;
}
}
